If the circles \(x^{2}+y^{2}-2 x-2 y-7=0\) and \(x^{2}+y^{2}+4 x+2 y+k=0\) cut orthogonally, then the length of the common chord of the circles is

Given circles are
\(
\begin{aligned}
&S_{1} \equiv x^{2}+y^{2}-2 x-2 y-7=0 \\
&S_{2} \equiv x^{2}+y^{2}+4 x+2 y+k=0
\end{aligned}
\)
Here, \(g_{1}=-1, f_{1}=-1, c_{1}=-7\)
\(
g_{2}=2, f_{2}=1, c_{2}=k
\)



Equation of common chord is \(S_{1}-S_{2}=0\)
\(
\begin{gathered}
\Rightarrow x^{2}+y^{2}-2 x-2 y-7-x^{2}-y^{2}-4 x \\
\Rightarrow \quad-2 y-k=0 \\
\Rightarrow \quad-6 x-4 y-7-k=0 \\
\end{gathered}
\)
Since, \(S_{1}\) and \(S_{2}\) cut orthogonally.
\(\begin{aligned} & \therefore & 2\left(g_{1} g_{2}+f_{1} f_{2}\right) &=c_{1}+c_{2} \\ \Rightarrow & 2(-2-1) &=-7+k \\ \Rightarrow & &-6+7 &=k \\ \Rightarrow & & k &=1 \\ & & & \\ & & & \end{aligned}\)
Now, length of the common chord
\(
r_{1}=\sqrt{1+1+7}=3
\)
\(c_{1}=(1,1)\)
Let \(C_{1} M=\) perpendicular distance from centre \(C_{1}(1,1)\) to the common chord
\(
\begin{aligned}
\Rightarrow C_{1} M &=\frac{|6+4+8|}{\sqrt{6^{2}+4^{2}}}=\frac{|18|}{\sqrt{5^{2}}}=\frac{18}{2 \sqrt{13}}=\frac{9}{\sqrt{13}} \\
\text { Now, } P Q &=2 P M=2 \sqrt{\left(C_{1} P\right)^{2}-\left(C_{1} M\right)^{2}} \\
&=2 \sqrt{9-\left(\frac{9}{\sqrt{13}}\right)^{2}} \\
&=2 \sqrt{9-\frac{81}{13}} \\
&=2 \sqrt{\frac{36}{13}}=\frac{12}{\sqrt{13}}
\end{aligned}
\)